4.7 Maxima/Minima Problems

One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.

Critical Points

For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Definition

Let z = f ( x , y ) z = f ( x , y ) be a function of two variables that is defined on an open set containing the point ( x 0 , y 0 ) . ( x 0 , y 0 ) . The point ( x 0 , y 0 ) ( x 0 , y 0 ) is called a critical point of a function of two variables f f if one of the two following conditions holds:

1. f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0 f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0
2. Either f x ( x 0 , y 0 ) or f y ( x 0 , y 0 ) f x ( x 0 , y 0 ) or f y ( x 0 , y 0 ) does not exist.

Example 4.38

Finding Critical Points

Find the critical points of each of the following functions:

1. f ( x , y ) = 4 y 2 − 9 x 2 + 24 y + 36 x + 36 f ( x , y ) = 4 y 2 − 9 x 2 + 24 y + 36 x + 36
2. g ( x , y ) = x 2 + 2 x y − 4 y 2 + 4 x − 6 y + 4 g ( x , y ) = x 2 + 2 x y − 4 y 2 + 4 x − 6 y + 4

Solution

1. First, we calculate f x ( x , y ) and f y ( x , y ) : f x ( x , y ) and f y ( x , y ) :
   

f x ( x , y ) = 1 2 ( −18 x + 36 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = −9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 f y ( x , y ) = 1 2 ( 8 y + 24 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 . f x ( x , y ) = 1 2 ( −18 x + 36 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = −9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 f y ( x , y ) = 1 2 ( 8 y + 24 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 .

−9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 . −9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 .

Therefore, x = 2 x = 2 and y = −3 , y = −3 , so ( 2 , −3 ) ( 2 , −3 ) is a critical point of f . f .
We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:

This equation represents a hyperbola. We should also note that the domain of f f consists of points satisfying the inequality

Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:

4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y 2 − 9 x 2 + 24 y + 36 x = −36 4 y 2 + 24 y − 9 x 2 + 36 x = −36 4 ( y 2 + 6 y ) − 9 ( x 2 − 4 x ) = −36 4 ( y 2 + 6 y + 9 ) − 9 ( x 2 − 4 x + 4 ) = −36 + 36 − 36 4 ( y + 3 ) 2 − 9 ( x − 2 ) 2 = −36 . 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y 2 − 9 x 2 + 24 y + 36 x = −36 4 y 2 + 24 y − 9 x 2 + 36 x = −36 4 ( y 2 + 6 y ) − 9 ( x 2 − 4 x ) = −36 4 ( y 2 + 6 y + 9 ) − 9 ( x 2 − 4 x + 4 ) = −36 + 36 − 36 4 ( y + 3 ) 2 − 9 ( x − 2 ) 2 = −36 .

4 ( y + 3 ) 2 −36 − 9 ( x − 2 ) 2 −36 = 1 ( x − 2 ) 2 4 − ( y + 3 ) 2 9 = 1 . 4 ( y + 3 ) 2 −36 − 9 ( x − 2 ) 2 −36 = 1 ( x − 2 ) 2 4 − ( y + 3 ) 2 9 = 1 .

g x ( x , y ) = 2 x + 2 y + 4 g y ( x , y ) = 2 x − 8 y − 6 . g x ( x , y ) = 2 x + 2 y + 4 g y ( x , y ) = 2 x − 8 y − 6 .

Next, we set each of these expressions equal to zero, which gives a system of equations in x and y : x and y :

Subtracting the second equation from the first gives 10 y + 10 = 0 , so y = −1 . 10 y + 10 = 0 , so y = −1 . Substituting this into the first equation gives 2 x + 2 ( −1 ) + 4 = 0 , 2 x + 2 ( −1 ) + 4 = 0 , so x = −1 . x = −1 . Therefore ( −1 , −1 ) ( −1 , −1 ) is a critical point of g g (Figure 4.46). There are no points in ℝ 2 ℝ 2 that make either partial derivative not exist.

Figure 4.46 The function g ( x , y ) g ( x , y ) has a critical point at ( −1 , −1 , 5 ) . ( −1 , −1 , 5 ) .

Checkpoint 4.34

Find the critical point of the function f ( x , y ) = x 3 + 2 x y − 2 x − 4 y . f ( x , y ) = x 3 + 2 x y − 2 x − 4 y .

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

Definition

Let z = f ( x , y ) z = f ( x , y ) be a function of two variables that is defined and continuous on an open set containing the point ( x 0 , y 0 ) . ( x 0 , y 0 ) . Then f has a local maximum at ( x 0 , y 0 ) ( x 0 , y 0 ) if

for all points ( x , y ) ( x , y ) within some disk centered at ( x 0 , y 0 ) . ( x 0 , y 0 ) . The number f ( x 0 , y 0 ) f ( x 0 , y 0 ) is called a local maximum value. If the preceding inequality holds for every point ( x , y ) ( x , y ) in the domain of f , f , then f f has a global maximum (also called an absolute maximum) at ( x 0 , y 0 ) . ( x 0 , y 0 ) .

The function f f has a local minimum at ( x 0 , y 0 ) ( x 0 , y 0 ) if

for all points ( x , y ) ( x , y ) within some disk centered at ( x 0 , y 0 ) . ( x 0 , y 0 ) . The number f ( x 0 , y 0 ) f ( x 0 , y 0 ) is called a local minimum value. If the preceding inequality holds for every point ( x , y ) ( x , y ) in the domain of f , f , then f f has a global minimum (also called an absolute minimum) at ( x 0 , y 0 ) . ( x 0 , y 0 ) .

If f ( x 0 , y 0 ) f ( x 0 , y 0 ) is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).

Figure 4.47 The graph of z = 16 − x 2 − y 2 z = 16 − x 2 − y 2 has a maximum value when ( x , y ) = ( 0 , 0 ) . ( x , y ) = ( 0 , 0 ) . It attains its minimum value at the boundary of its domain, which is the circle x 2 + y 2 = 16 . x 2 + y 2 = 16 .

In Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Theorem 4.16

Fermat’s Theorem for Functions of Two Variables

Let z = f ( x , y ) z = f ( x , y ) be a function of two variables that is defined and continuous on an open set containing the point ( x 0 , y 0 ) . ( x 0 , y 0 ) . Suppose f x f x and f y f y each exists at ( x 0 , y 0 ) . ( x 0 , y 0 ) . If f f has a local extremum at ( x 0 , y 0 ) , ( x 0 , y 0 ) , then ( x 0 , y 0 ) ( x 0 , y 0 ) is a critical point of f . f .

Second Derivative Test

Consider the function f ( x ) = x 3 . f ( x ) = x 3 . This function has a critical point at x = 0 , x = 0 , since f ′ ( 0 ) = 3 ( 0 ) 2 = 0 . f ′ ( 0 ) = 3 ( 0 ) 2 = 0 . However, f f does not have an extreme value at x = 0 . x = 0 . Therefore, the existence of a critical value at x = x 0 x = x 0 does not guarantee a local extremum at x = x 0 . x = x 0 . The same is true for a function of two or more variables. One way this can happen is at a saddle point . An example of a saddle point appears in the following figure.

Figure 4.48 Graph of the function z = x 2 − y 2 . z = x 2 − y 2 . This graph has a saddle point at the origin.

In this graph, the origin is a saddle point. This is because the first partial derivatives of f ( x , y ) = x 2 − y 2 f ( x , y ) = x 2 − y 2 are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to y = 0 y = 0 is z = x 2 z = x 2 (a parabola opening upward), but the vertical trace corresponding to x = 0 x = 0 is z = − y 2 z = − y 2 (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

Definition

Given the function z = f ( x , y ) , z = f ( x , y ) , the point ( x 0 , y 0 , f ( x 0 , y 0 ) ) ( x 0 , y 0 , f ( x 0 , y 0 ) ) is a saddle point if both f x ( x 0 , y 0 ) = 0 f x ( x 0 , y 0 ) = 0 and f y ( x 0 , y 0 ) = 0 , f y ( x 0 , y 0 ) = 0 , but f f does not have a local extremum at ( x 0 , y 0 ) . ( x 0 , y 0 ) .

The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant D D that replaces f ″ ( x 0 ) f ″ ( x 0 ) in the second derivative test for a function of one variable.

Theorem 4.17

Second Derivative Test

Let z = f ( x , y ) z = f ( x , y ) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ( x 0 , y 0 ) . ( x 0 , y 0 ) . Suppose f x ( x 0 , y 0 ) = 0 f x ( x 0 , y 0 ) = 0 and f y ( x 0 , y 0 ) = 0 . f y ( x 0 , y 0 ) = 0 . Define the quantity

D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) − ( f x y ( x 0 , y 0 ) ) 2 . D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) − ( f x y ( x 0 , y 0 ) ) 2 .

1. If D > 0 D > 0 and f x x ( x 0 , y 0 ) > 0 , f x x ( x 0 , y 0 ) > 0 , then f f has a local minimum at ( x 0 , y 0 ) . ( x 0 , y 0 ) .
2. If D > 0 D > 0 and f x x ( x 0 , y 0 ) < 0 , f x x ( x 0 , y 0 ) < 0 , then f f has a local maximum at ( x 0 , y 0 ) . ( x 0 , y 0 ) .
3. If D < 0 , D < 0 , then f f has a saddle point at ( x 0 , y 0 ) . ( x 0 , y 0 ) .
4. If D = 0 , D = 0 , then the test is inconclusive.

Figure 4.49 The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).

To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.

Problem-Solving Strategy

Using the Second Derivative Test for Functions of Two Variables

Let z = f ( x , y ) z = f ( x , y ) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ( x 0 , y 0 ) . ( x 0 , y 0 ) . To apply the second derivative test to find local extrema, use the following steps:

1. Determine the critical points ( x 0 , y 0 ) ( x 0 , y 0 ) of the function f f where f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0 . f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0 . Discard any points where at least one of the partial derivatives does not exist.
2. Calculate the discriminant D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) − ( f x y ( x 0 , y 0 ) ) 2 D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) − ( f x y ( x 0 , y 0 ) ) 2 for each critical point of f . f .
3. Apply Second Derivative Test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.

Example 4.39

Using the Second Derivative Test

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

1. f ( x , y ) = 4 x 2 + 9 y 2 + 8 x − 36 y + 24 f ( x , y ) = 4 x 2 + 9 y 2 + 8 x − 36 y + 24
2. g ( x , y ) = 1 3 x 3 + y 2 + 2 x y − 6 x − 3 y + 4 g ( x , y ) = 1 3 x 3 + y 2 + 2 x y − 6 x − 3 y + 4

Solution

1. Step 1 1 of the problem-solving strategy involves finding the critical points of f . f . To do this, we first calculate f x ( x , y ) f x ( x , y ) and f y ( x , y ) , f y ( x , y ) , then set each of them equal to zero:
   

f x ( x , y ) = 8 x + 8 f y ( x , y ) = 18 y − 36 . f x ( x , y ) = 8 x + 8 f y ( x , y ) = 18 y − 36 .

The solution to this system is x = −1 x = −1 and y = 2 . y = 2 . Therefore ( −1 , 2 ) ( −1 , 2 ) is a critical point of f . f .
Step 2 of the problem-solving strategy involves calculating D . D . To do this, we first calculate the second partial derivatives of f : f :

f x x ( x , y ) = 8 f x y ( x , y ) = 0 f y y ( x , y ) = 18 . f x x ( x , y ) = 8 f x y ( x , y ) = 0 f y y ( x , y ) = 18 .

Therefore, D = f x x ( −1 , 2 ) f y y ( −1 , 2 ) − ( f x y ( −1 , 2 ) ) 2 = ( 8 ) ( 18 ) − ( 0 ) 2 = 144 . D = f x x ( −1 , 2 ) f y y ( −1 , 2 ) − ( f x y ( −1 , 2 ) ) 2 = ( 8 ) ( 18 ) − ( 0 ) 2 = 144 .
Step 3 states to check the Second Derivative Test for Functions of Two Variables.. Since D > 0 D > 0 and f x x ( −1 , 2 ) > 0 , f x x ( −1 , 2 ) > 0 , this corresponds to case 1. Therefore, f f has a local minimum at ( −1 , 2 ) ( −1 , 2 ) as shown in the following figure.

Figure 4.50 The function f ( x , y ) f ( x , y ) has a local minimum at ( −1 , 2 , −16 ) . ( −1 , 2 , −16 ) .

g x ( x , y ) = x 2 + 2 y − 6 g y ( x , y ) = 2 y + 2 x − 3 . g x ( x , y ) = x 2 + 2 y − 6 g y ( x , y ) = 2 y + 2 x − 3 .

To solve this system, first solve the second equation for y. This gives y = 3 − 2 x 2 . y = 3 − 2 x 2 . Substituting this into the first equation gives

x 2 + 3 − 2 x − 6 = 0 x 2 − 2 x − 3 = 0 ( x − 3 ) ( x + 1 ) = 0 . x 2 + 3 − 2 x − 6 = 0 x 2 − 2 x − 3 = 0 ( x − 3 ) ( x + 1 ) = 0 .

Therefore, x = −1 x = −1 or x = 3 . x = 3 . Substituting these values into the equation y = 3 − 2 x 2 y = 3 − 2 x 2 yields the critical points ( −1 , 5 2 ) ( −1 , 5 2 ) and ( 3 , − 3 2 ) . ( 3 , − 3 2 ) . Step 2 involves calculating the second partial derivatives of g : g :

g x x ( x , y ) = 2 x g x y ( x , y ) = 2 g y y ( x , y ) = 2 . g x x ( x , y ) = 2 x g x y ( x , y ) = 2 g y y ( x , y ) = 2 .

D = g x x ( x 0 , y 0 ) g y y ( x 0 , y 0 ) − ( g x y ( x 0 , y 0 ) ) 2 = ( 2 x 0 ) ( 2 ) − 2 2 = 4 x 0 − 4 . D = g x x ( x 0 , y 0 ) g y y ( x 0 , y 0 ) − ( g x y ( x 0 , y 0 ) ) 2 = ( 2 x 0 ) ( 2 ) − 2 2 = 4 x 0 − 4 .

D ( −1 , 5 2 ) = ( 2 ( −1 ) ) ( 2 ) − ( 2 ) 2 = −4 − 4 = −8 D ( 3 , − 3 2 ) = ( 2 ( 3 ) ) ( 2 ) − ( 2 ) 2 = 12 − 4 = 8 . D ( −1 , 5 2 ) = ( 2 ( −1 ) ) ( 2 ) − ( 2 ) 2 = −4 − 4 = −8 D ( 3 , − 3 2 ) = ( 2 ( 3 ) ) ( 2 ) − ( 2 ) 2 = 12 − 4 = 8 .

In step 3, we note that, applying the Second Derivative Test for Functions of Two Variables. to point ( −1 , 5 2 ) ( −1 , 5 2 ) leads to case 3 , 3 , which means that ( −1 , 5 2 ) ( −1 , 5 2 ) is a saddle point. Applying the theorem to point ( 3 , − 3 2 ) ( 3 , − 3 2 ) leads to case 1, which means that ( 3 , − 3 2 ) ( 3 , − 3 2 ) corresponds to a local minimum as shown in the following figure.

Checkpoint 4.35

Use the second derivative to find the local extrema of the function

Absolute Maxima and Minima

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

Theorem 4.18

Extreme Value Theorem

A continuous function f ( x , y ) f ( x , y ) on a closed and bounded set D D in the plane attains an absolute maximum value at some point of D D and an absolute minimum value at some point of D . D .

Now that we know any continuous function f f defined on a closed, bounded set attains its extreme values, we need to know how to find them.

Theorem 4.19

Finding Extreme Values of a Function of Two Variables

Assume z = f ( x , y ) z = f ( x , y ) is a differentiable function of two variables defined on a closed, bounded set D . D . Then f f will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

1. The values of f f at the critical points of f f in D . D .
2. The values of f f on the boundary of D . D .

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of D , D , then it is located at an interior point of D . D . But an interior point ( x 0 , y 0 ) ( x 0 , y 0 ) of D D that’s an absolute extremum is also a local extremum; hence, ( x 0 , y 0 ) ( x 0 , y 0 ) is a critical point of f f by Fermat’s theorem. Therefore the only possible values for the global extrema of f f on D D are the extreme values of f f on the interior or boundary of D . D .

Problem-Solving Strategy

Finding Absolute Maximum and Minimum Values

Let z = f ( x , y ) z = f ( x , y ) be a continuous function of two variables defined on a closed, bounded set D , D , and assume f f is differentiable on D . D . To find the absolute maximum and minimum values of f f on D , D , do the following:

1. Determine the critical points of f f in D . D .
2. Calculate f f at each of these critical points.
3. Determine the maximum and minimum values of f f on the boundary of its domain.
4. The maximum and minimum values of f f will occur at one of the values obtained in steps 2 and 3 . 2 and 3 .

Finding the maximum and minimum values of f f on the boundary of D D can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example 4.40. The same approach can be used for other shapes such as circles and ellipses.

If the boundary of the set D D is a more complicated curve defined by a function g ( x , y ) = c g ( x , y ) = c for some constant c , c , and the first-order partial derivatives of g g exist, then the method of Lagrange multipliers can prove useful for determining the extrema of f f on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers.

Example 4.40

Finding Absolute Extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

1. f ( x , y ) = x 2 − 2 x y + 4 y 2 − 4 x − 2 y + 24 f ( x , y ) = x 2 − 2 x y + 4 y 2 − 4 x − 2 y + 24 on the domain defined by 0 ≤ x ≤ 4 0 ≤ x ≤ 4 and 0 ≤ y ≤ 2 0 ≤ y ≤ 2
2. g ( x , y ) = x 2 + y 2 + 4 x − 6 y g ( x , y ) = x 2 + y 2 + 4 x − 6 y on the domain defined by x 2 + y 2 ≤ 16 x 2 + y 2 ≤ 16

Solution

1. Using the problem-solving strategy, step 1 1 involves finding the critical points of f f on its domain. Therefore, we first calculate f x ( x , y ) f x ( x , y ) and f y ( x , y ) , f y ( x , y ) , then set them each equal to zero:
   

f x ( x , y ) = 2 x − 2 y − 4 f y ( x , y ) = −2 x + 8 y − 2 . f x ( x , y ) = 2 x − 2 y − 4 f y ( x , y ) = −2 x + 8 y − 2 .

The solution to this system is x = 3 x = 3 and y = 1 . y = 1 . Therefore ( 3 , 1 ) ( 3 , 1 ) is a critical point of f . f . Calculating f ( 3 , 1 ) f ( 3 , 1 ) gives f ( 3 , 1 ) = 17 . f ( 3 , 1 ) = 17 .
The next step involves finding the extrema of f f on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:

Figure 4.52 Graph of the domain of the function f ( x , y ) = x 2 − 2 x y + 4 y 2 − 4 x − 2 y + 24 . f ( x , y ) = x 2 − 2 x y + 4 y 2 − 4 x − 2 y + 24 .

L 1 L 1 is the line segment connecting ( 0 , 0 ) ( 0 , 0 ) and ( 4 , 0 ) , ( 4 , 0 ) , and it can be parameterized by the equations x ( t ) = t , y ( t ) = 0 x ( t ) = t , y ( t ) = 0 for 0 ≤ t ≤ 4 . 0 ≤ t ≤ 4 . Define g ( t ) = f ( x ( t ) , y ( t ) ) . g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = t 2 − 4 t + 24 . g ( t ) = t 2 − 4 t + 24 . Differentiating g leads to g ′ ( t ) = 2 t − 4 . g ′ ( t ) = 2 t − 4 . Therefore, g g has a critical value at t = 2 , t = 2 , which corresponds to the point ( 2 , 0 ) . ( 2 , 0 ) . Calculating f ( 2 , 0 ) f ( 2 , 0 ) gives the z-value 20 . 20 .
L 2 L 2 is the line segment connecting ( 4 , 0 ) ( 4 , 0 ) and ( 4 , 2 ) , ( 4 , 2 ) , and it can be parameterized by the equations x ( t ) = 4 , y ( t ) = t x ( t ) = 4 , y ( t ) = t for 0 ≤ t ≤ 2 . 0 ≤ t ≤ 2 . Again, define g ( t ) = f ( x ( t ) , y ( t ) ) . g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = 4 t 2 − 10 t + 24 . g ( t ) = 4 t 2 − 10 t + 24 . Then, g ′ ( t ) = 8 t − 10 . g ′ ( t ) = 8 t − 10 . g g has a critical value at t = 5 4 , t = 5 4 , which corresponds to the point ( 4 , 5 4 ) . ( 4 , 5 4 ) . Calculating f ( 4 , 5 4 ) f ( 4 , 5 4 ) gives the z-value 17.75 . 17.75 .
L 3 L 3 is the line segment connecting ( 0 , 2 ) ( 0 , 2 ) and ( 4 , 2 ) , ( 4 , 2 ) , and it can be parameterized by the equations x ( t ) = t , y ( t ) = 2 x ( t ) = t , y ( t ) = 2 for 0 ≤ t ≤ 4 . 0 ≤ t ≤ 4 . Again, define g ( t ) = f ( x ( t ) , y ( t ) ) . g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = t 2 − 8 t + 36 . g ( t ) = t 2 − 8 t + 36 . The critical value corresponds to the point ( 4 , 2 ) . ( 4 , 2 ) . So, calculating f ( 4 , 2 ) f ( 4 , 2 ) gives the z-value 20 . 20 .
L 4 L 4 is the line segment connecting ( 0 , 0 ) ( 0 , 0 ) and ( 0 , 2 ) , ( 0 , 2 ) , and it can be parameterized by the equations x ( t ) = 0 , y ( t ) = t x ( t ) = 0 , y ( t ) = t for 0 ≤ t ≤ 2 . 0 ≤ t ≤ 2 . This time, g ( t ) = 4 t 2 − 2 t + 24 g ( t ) = 4 t 2 − 2 t + 24 and the critical value t = 1 4 t = 1 4 correspond to the point ( 0 , 1 4 ) . ( 0 , 1 4 ) . Calculating f ( 0 , 1 4 ) f ( 0 , 1 4 ) gives the z-value 23.75 . 23.75 .
We also need to find the values of f ( x , y ) f ( x , y ) at the corners of its domain. These corners are located at ( 0 , 0 ) , ( 4 , 0 ) , ( 4 , 2 ) and ( 0 , 2 ) : ( 0 , 0 ) , ( 4 , 0 ) , ( 4 , 2 ) and ( 0 , 2 ) :

f ( 0 , 0 ) = ( 0 ) 2 − 2 ( 0 ) ( 0 ) + 4 ( 0 ) 2 − 4 ( 0 ) − 2 ( 0 ) + 24 = 24 f ( 4 , 0 ) = ( 4 ) 2 − 2 ( 4 ) ( 0 ) + 4 ( 0 ) 2 − 4 ( 4 ) − 2 ( 0 ) + 24 = 24 f ( 4 , 2 ) = ( 4 ) 2 − 2 ( 4 ) ( 2 ) + 4 ( 2 ) 2 − 4 ( 4 ) − 2 ( 2 ) + 24 = 20 f ( 0 , 2 ) = ( 0 ) 2 − 2 ( 0 ) ( 2 ) + 4 ( 2 ) 2 − 4 ( 0 ) − 2 ( 2 ) + 24 = 36 . f ( 0 , 0 ) = ( 0 ) 2 − 2 ( 0 ) ( 0 ) + 4 ( 0 ) 2 − 4 ( 0 ) − 2 ( 0 ) + 24 = 24 f ( 4 , 0 ) = ( 4 ) 2 − 2 ( 4 ) ( 0 ) + 4 ( 0 ) 2 − 4 ( 4 ) − 2 ( 0 ) + 24 = 24 f ( 4 , 2 ) = ( 4 ) 2 − 2 ( 4 ) ( 2 ) + 4 ( 2 ) 2 − 4 ( 4 ) − 2 ( 2 ) + 24 = 20 f ( 0 , 2 ) = ( 0 ) 2 − 2 ( 0 ) ( 2 ) + 4 ( 2 ) 2 − 4 ( 0 ) − 2 ( 2 ) + 24 = 36 .

The absolute maximum value is 36 , 36 , which occurs at ( 0 , 2 ) , ( 0 , 2 ) , and the global minimum value is 17 , 17 , which occurs at ( 3 , 1 ) ( 3 , 1 ) as shown in the following figure.

Figure 4.53 The function f ( x , y ) f ( x , y ) has one global minimum and one global maximum over its domain.

The solution to this system is x = −2 x = −2 and y = 3 . y = 3 . Therefore, ( −2 , 3 ) ( −2 , 3 ) is a critical point of g . g . Calculating g ( −2 , 3 ) , g ( −2 , 3 ) , we get

g ( −2 , 3 ) = ( −2 ) 2 + 3 2 + 4 ( −2 ) − 6 ( 3 ) = 4 + 9 − 8 − 18 = −13 . g ( −2 , 3 ) = ( −2 ) 2 + 3 2 + 4 ( −2 ) − 6 ( 3 ) = 4 + 9 − 8 − 18 = −13 .

The next step involves finding the extrema of g on the boundary of its domain. The boundary of its domain consists of a circle of radius 4 4 centered at the origin as shown in the following graph.

Figure 4.54 Graph of the domain of the function g ( x , y ) = x 2 + y 2 + 4 x − 6 y . g ( x , y ) = x 2 + y 2 + 4 x − 6 y .

The boundary of the domain of g g can be parameterized using the functions x ( t ) = 4 cos t , y ( t ) = 4 sin t x ( t ) = 4 cos t , y ( t ) = 4 sin t for 0 ≤ t ≤ 2 π . 0 ≤ t ≤ 2 π . Define h ( t ) = g ( x ( t ) , y ( t ) ) : h ( t ) = g ( x ( t ) , y ( t ) ) :

h ( t ) = g ( x ( t ) , y ( t ) ) = ( 4 cos t ) 2 + ( 4 sin t ) 2 + 4 ( 4 cos t ) − 6 ( 4 sin t ) = 16 cos 2 t + 16 sin 2 t + 16 cos t − 24 sin t = 16 + 16 cos t − 24 sin t . h ( t ) = g ( x ( t ) , y ( t ) ) = ( 4 cos t ) 2 + ( 4 sin t ) 2 + 4 ( 4 cos t ) − 6 ( 4 sin t ) = 16 cos 2 t + 16 sin 2 t + 16 cos t − 24 sin t = 16 + 16 cos t − 24 sin t .

− 16 sin t − 24 cos t = 0 − 16 sin t = 24 cos t −16 sin t −16 cos t = 24 cos t −16 cos t tan t = − 3 2 . − 16 sin t − 24 cos t = 0 − 16 sin t = 24 cos t −16 sin t −16 cos t = 24 cos t −16 cos t tan t = − 3 2 .

This equation has two solutions over the interval 0 ≤ t ≤ 2 π . 0 ≤ t ≤ 2 π . One is t = π − arctan ( 3 2 ) t = π − arctan ( 3 2 ) and the other is t = 2 π − arctan ( 3 2 ) . t = 2 π − arctan ( 3 2 ) . For the first angle,

sin t = sin ( π − arctan ( 3 2 ) ) = sin ( arctan ( 3 2 ) ) = 3 13 13 cos t = cos ( π − arctan ( 3 2 ) ) = − cos ( arctan ( 3 2 ) ) = − 2 13 13 . sin t = sin ( π − arctan ( 3 2 ) ) = sin ( arctan ( 3 2 ) ) = 3 13 13 cos t = cos ( π − arctan ( 3 2 ) ) = − cos ( arctan ( 3 2 ) ) = − 2 13 13 .

Therefore, x ( t ) = 4 cos t = − 8 13 13 x ( t ) = 4 cos t = − 8 13 13 and y ( t ) = 4 sin t = 12 13 13 , y ( t ) = 4 sin t = 12 13 13 , so ( − 8 13 13 , 12 13 13 ) ( − 8 13 13 , 12 13 13 ) is a critical point on the boundary and

g ( − 8 13 13 , 12 13 13 ) = ( − 8 13 13 ) 2 + ( 12 13 13 ) 2 + 4 ( − 8 13 13 ) − 6 ( 12 13 13 ) = 144 13 + 64 13 − 32 13 13 − 72 13 13 = 208 − 104 13 13 ≈ −12.844 . g ( − 8 13 13 , 12 13 13 ) = ( − 8 13 13 ) 2 + ( 12 13 13 ) 2 + 4 ( − 8 13 13 ) − 6 ( 12 13 13 ) = 144 13 + 64 13 − 32 13 13 − 72 13 13 = 208 − 104 13 13 ≈ −12.844 .

sin t = sin ( 2 π − arctan ( 3 2 ) ) = − sin ( arctan ( 3 2 ) ) = − 3 13 13 cos t = cos ( 2 π − arctan ( 3 2 ) ) = cos ( arctan ( 3 2 ) ) = 2 13 13 . sin t = sin ( 2 π − arctan ( 3 2 ) ) = − sin ( arctan ( 3 2 ) ) = − 3 13 13 cos t = cos ( 2 π − arctan ( 3 2 ) ) = cos ( arctan ( 3 2 ) ) = 2 13 13 .

Therefore, x ( t ) = 4 cos t = 8 13 13 x ( t ) = 4 cos t = 8 13 13 and y ( t ) = 4 sin t = − 12 13 13 , y ( t ) = 4 sin t = − 12 13 13 , so ( 8 13 13 , − 12 13 13 ) ( 8 13 13 , − 12 13 13 ) is a critical point on the boundary and

g ( 8 13 13 , − 12 13 13 ) = ( 8 13 13 ) 2 + ( − 12 13 13 ) 2 + 4 ( 8 13 13 ) − 6 ( − 12 13 13 ) = 144 13 + 64 13 + 32 13 13 + 72 13 13 = 208 + 104 13 13 ≈ 44.844 . g ( 8 13 13 , − 12 13 13 ) = ( 8 13 13 ) 2 + ( − 12 13 13 ) 2 + 4 ( 8 13 13 ) − 6 ( − 12 13 13 ) = 144 13 + 64 13 + 32 13 13 + 72 13 13 = 208 + 104 13 13 ≈ 44.844 .

The absolute minimum of g is −13 , −13 , which is attained at the point ( −2 , 3 ) , ( −2 , 3 ) , which is an interior point of D. The absolute maximum of g is approximately equal to 44.844, which is attained at the boundary point ( 8 13 13 , − 12 13 13 ) . ( 8 13 13 , − 12 13 13 ) . These are the absolute extrema of g on D as shown in the following figure.